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In-s [12.5K]
3 years ago
10

Change the fraction 13/20 to a decimal. A. 0.65 B. 0.015 C. 0.065 D. 0.15

Mathematics
2 answers:
Serga [27]3 years ago
7 0
The answer is A

set 13/20=x/100 13x100 is 1300 divided by 20 is 65.
65/100 is 0.65 as a decimal
Anastaziya [24]3 years ago
4 0

Answer:

A is going to be correct

Step-by-step explanation:


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Meg is 6 years older than Victor. Meg's age is 2 years less than five times Victor's age. The equations below model the relation
Verdich [7]

Since no possible correct method is posted, I will suggest a couple.

Method 1: guess and check

Works well for simple problems involving integers like this one.

Victor's age must be zero or greater than one, say one.

Guess v=1, find m=v+6=7, check m=5v-2=5-2=3 no good.

we need to make v bigger

Guess v=2, find m=v+6=2+6=8, check m=5v-2=5*2-2=8 ✔

So v=2, m=8.

Method 2:

Solve the system of two equations.

since the left-hand sides is m in both equations, and since m=m, we just have to equate the right-hand sides to solve for v.

5v-2=v+6

Solve for v

5v-v = 6+2

4v=8

v=2,

so again, v=2, m=v+6=2+6=8.

6 0
4 years ago
HELPPP MEEE PLZZZ AND THANKS
Dmitry [639]
-3
Explanation - you add the 10 to both sides then get -12 and then you divide 4 to both sides and then get -3
3 0
3 years ago
Read 2 more answers
Given that BE=6, CE=6,<br>and FE measures 5 more<br>than DE, find the length<br>of DE.​
katen-ka-za [31]

Answer:

DE = 4 units.

Step-by-step explanation:

By the chord intersection theorem:

BE * CE = DE * FE

But FE = DE + 5 so:

BE * CE = DE * (DE + 5)

6 * 6 = DE^2 + 5DE

DE^2 + 5DE  - 36 = 0

(DE + 9)(DE - 4) = 0

So DE = -9 or 4

It can't be negative so DE = 4.

4 0
4 years ago
Find the percent of change when 88 is decreased to 73.
gtnhenbr [62]
Its 
-17.0455
v2-v1
----------
v1
=
73-88
--------- x100
88    
=
-15
 -----     x100
88 
answer:
17.0455

8 0
3 years ago
Find the derivative of<br> <img src="https://tex.z-dn.net/?f=f%28x%29%20%3D%20%20%5Cfrac%7B6%7D%7Bx%7D%20" id="TexFormula1" titl
aev [14]
f'(x_0)=\lim\limits_{h\to0}\dfrac{f(x_0+h)-f(h)}{h}=\lim\limits_{x\to x_0}\dfrac{f(x)-f(x_0)}{x-x_0}\\\\f(x)=\dfrac{6}{x};\ x_0=2\\\\subtitute\\\\f'(2)=\lim\limits_{x\to2}\dfrac{\frac{6}{x}-\frac{6}{2}}{x-2}=\lim\limits_{x\to2}\dfrac{\frac{6}{x}-3}{x-2}=\lim\limits_{x\to2}\dfrac{\frac{6-3x}{x}}{x-2}=\lim\limits_{x\to 2}\dfrac{6-3x}{x(x-2)}\\\\=\lim\limits_{x\to2}\dfrac{-3(x-2)}{x(x-2)}=\lim\limits_{x\to2}\dfrac{-3}{x}=-\dfrac{3}{2}=-1.5\\\\\\An swer:\boxed{f'(2)=-1.5}
8 0
3 years ago
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